Question: Let $f(x) = 10x^{2}+4x-8$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )?
The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $10x^{2}+4x-8 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 10, b = 4, c = -8$ $ x = \dfrac{-4 \pm \sqrt{4^{2} - 4 \cdot 10 \cdot -8}}{2 \cdot 10}$ $ x = \dfrac{-4 \pm \sqrt{336}}{20}$ $ x = \dfrac{-4 \pm 4\sqrt{21}}{20}$ $x =\dfrac{-1 \pm \sqrt{21}}{5}$